One-dimensional transient heat conduction in finite slab

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For the case that the Biot number is greater than 0.1, the temperature distribution can no longer be treated as uniform and the knowledge about the temperature distribution is also of interest. We will now consider the situation where temperature only varies in one spatial dimension. Both homogeneous and nonhomogeneous problems will be considered.

Contents

Homogeneous Problems

Figure 1 Transient conduction in a finite slab

Figure 1 shows a finite slab with thickness of L and a uniform initial temperature of Ti. At time t= 0, the left side of the slab is insulated while the right side of the slab is exposed to a fluid with temperature of (). The convective heat transfer coefficient between the fluid and finite slab is h. In contrast to the lumped capacitance method that assumes uniform temperature, we will present a more generalized model that takes non-uniform temperature distribution in the slab into account. It is assumed that there is no internal heat generation in the slab.
The energy equation for this one-dimensional transient conduction problem is

subject to the following boundary and initial conditions

It should be pointed out that eqs. (1) – (4) are also valid for the case that both sides of a finite slab with thickness of 2L are cooled by convection. This is a nonhomogeneous problem because eq. (3) is not homogeneous. By introducing the excess temperature, , the problem can be homogenized, i.e.,

To express our solution in a compact form so that it can be used for all similar problems, one can define the following dimensionless variables

and eqs. (5) – (8) will be nondimensionalized as

This problem can be solved using the method of separation of variables. Assuming that the temperature can be expressed as

where ΘandΓ are functions of X and Fo, respectively, eq. (10) becomes

Since the left hand side is a function of X only and the right-hand side of the above equation is a function of Fo only, both sides must be equal to a separation constant, μ, i.e.,

The separation constant μ can be either a real or a complex number. The solution of Γ from eq. (15) will be Γ = eμFo. If μ is a positive real number, we will have when , which does not make sense, therefore, μ cannot be a positive real number. If μ is zero, we will have Γ = const, and Θ is a linear function of X only. The final solution for θ will also be a linear function of X only, which does not make sense either. It can also be shown that the separation constant cannot be a complex number (see Problem 3.24), therefore, the separation variable has to be a negative real number. If we represent this negative number by μ = − λ2, eq. (15) can be rewritten as the following two equations

The general solutions of eqs. (16) and (17) are

where C1,C2, and C3 are integral constants.
Substituting eq. (14) into eqs. (11) and (12), the following boundary conditions of eq. (16) are obtained

Substituting eq. (18) into eq. (20) yields

Θ'(0) = − C1λsin(0) + C2λcos(0) = C2λ = 0

Since λ cannot be zero, C2 must be zero and eq. (18) becomes

Applying the convection boundary condition, eq. (21), one obtains

where n is an integer. Equation (23) indicates that many possible values for λ – termed eigenvalues – can satisfy the convection condition. The eigenvalue λn can be obtained by solving eq. (23) using an iterative procedure. The dimensionless temperature with eigenvalue λn can be obtained by substituting eqs. (22) and (19) into eq. (14), i.e.,

where Cn = C1C3. Equation (24) is a solution that satisfies eqs. (10) – (12). At this point, the constant Cn is still unspecified and eq. (13) is unused, however, if we substitute eq. (24) into (13), the constant Cn that satisfies the initial condition cannot be found. Since the one-dimensional transient heat conduction problem under consideration is a linear problem, the sum of different θn for each value of n also satisfies eqs. (10) – (12).

Substituting eq. (25) into eq. (13) yields

Multiplying the above equation by and integrating the resulting equation in the interval of (0, 1), one obtains

The integral in the right-hand side of eq. (26) can be evaluated as

Equation (23) can be rewritten as

Bi = λntanλn

Similarly, for eigenvalue λm, we have

Bi = λmtanλm

Combining the above two equations, we have

λmtanλm − λntanλn = 0

or

λmsinλmcosλn − λncosλmsinλn = 0

therefore, the integral in eq. (27) is zero for the case that , and the right hand side of eq. (26) becomes

Substituting eq. (28) into eq. (26) and evaluating the integral at the left-hand side of eq. (26), we have

i.e.,

Changing notation from m to n, we get

The dimensionless temperature, therefore, becomes

If the Biot number becomes infinite, the convection boundary condition becomes

which is an isothermal condition at the right-hand side of the wall. Equation (23) becomes

and the eigenvalue is therefore

The temperature distribution for this case is then

When Fourier’s number is greater than 0.2, only the first term in eq. (25) is necessary and the solution becomes

Nonhomogeneous Problems

The transient one-dimensional conduction problems that we discussed so far are limited to the case that the problem is homogeneous and the method of separation of variables works. When the problem is not homogeneous due to a nonhomogeneous energy equation or boundary condition, the solution of a nonhomogeneous problem can be obtained by superposition of a particular solution of the nonhomogeneous problem and the general solution of the corresponding homogeneous problem.

Partial Solution

Figure 2: Heat conduction under boundary condition of the first kind

Let us consider a finite slab with thickness of L and a uniform initial temperature of Ti as shown in the figure on the right. At time t= 0, the temperature on the left side of the slab is suddenly increased to T0 while the temperature on the right side of the slab is maintained at Ti. Assuming that there is no internal heat generation in the slab and the thermophysical properties of the slab are constants, the energy equation is

subject to the following boundary and initial conditions

By defining the following dimensionless variables

eqs. (36) – (39) will be nondimensionalized as

It can be seen that the problem is still nonhomogeneous after nondimensionalization because eq. (42) is not homogeneous. When (i.e., ), the temperature distribution can reach to steady state. If the steady state temperature is represented by θs, it must satisfy the following equations:

which have the following solution:

To obtain the generation of the problem described by eqs. (41) – (44), a method of partial solution (Myers, 1987) will be employed. In this methodology, it is assumed that the solution of a nonhomogeneous problem can be expressed as

where θh represent the solution of a homogeneous problem. Substituting eqs. (41) – (44) and considering eqs. (45) – (47), we have

which represent a new homogeneous problem. This problem can be solved using the method of separation of variables (see Problem 3.31) and the result is

The solution of the nonhomogeneous problem thus becomes

Variation of Parameter

Figure 3: Heat conduction under boundary condition of the second kind

The partial solution only works if the steady-state solution exists. If the steady-state solution does not exist, we can use the method of variation of parameters to solve the problem. Let us consider a finite slab with thickness of L and a uniform initial temperature of Ti. At time t= 0, the left side is subject to a constant heat flux while the right side of the slab is adiabatic (see Fig. 3). Assuming that there is no internal heat generation in the slab and the thermophysical properties of the slab are constants, the energy equation is

subject to the following boundary and initial conditions

By defining the following dimensionless variables

eqs. (56) – (59) will be nondimensionalized as

This nonhomogeneous problem does not have a steady-state solution, and therefore the partial solution cannot be applied. We will use the method of variation of parameters (Myers, 1987) to solve this problem. This method requires the following steps:

1. Set up a homogeneous problem by dropping the nonhomogeneous terms.

2. Solve the homogeneous problem to get eigenvalue λn and eigenfunctions Θn(X)

3. Assuming the solution of the original nonhomogeneous problem has the form of

4. Solve for An(Fo) using orthogonal property of Θn

5. Obtain an ordinary differential equation (ODE) for An(Fo) and solve for An(Fo) from the ODE

6. Put together the final solution.

We will solve this nonhomogeneous problem by following the above procedure.
The corresponding homogeneous problem is:

Assuming the solution of the above homogeneous problem is

eq. (65) becomes

Since the objective here is to get the eigenvalue and eigen functions, we do not need to solve for Γ and only need to solve for Θ. The eigenvalue problem is

Now, let us assume that the solution of the original nonhomogeneous problem is

Multiplying eq. (76) by and integrating the resulting equation in the interval of (0, 1), one obtains

The integral on the right-hand side of eq. (77) can be evaluated as

thus, eq. (77) becomes

Differentiating eq. (79) with respect to Fo, one obtains:

Substituting eq. (61) into eq. (81) and integrating with respect to X yields

Integrating eq. (82) with respect to Fo, we have

When Fo = 0, eq. (83) becomes

thus, we have

Differentiating eq. (80) and considering eq. (61) yield

Using integration by parts twice, the following ODE is obtained:

Multiplying eq. (87) by an integrating factor , we have

which can be integrated to get

where C2 is an integral constant that needs to be determined by an initial condition. For Fo = 0, eq. (80) becomes

Substituting eq. (89) into eq. (90), one obtains

therefore, we have

Changing m back to n for notation,

Substituting eqs. (85) and (91) into eq. (76), the solution becomes

When the time (Fourier number) becomes large, the last term on the right-hand side will become zero and the solution is represented by the first two terms only. To simplify eq. (92), let us assume the solution at large Fo can be expressed as

which is referred to as an asymptotic solution and it must satisfy eqs. (61) – (63). Substituting eq. (93) into eqs. (61) – (63), we have

Integrating eq. (94) and considering eq. (95), we obtain

where C cannot be determined from eq. (95) because both boundary conditions are for the first order derivative. To determine C, we can expand f(X) defined in eq. (96) into cosine Fourier series, i.e.

After determining a0 and an, and considering that f(X) is identical to the second term on the right-hand side of eq. (92), we have